# Numbers of form 31 x 16^n are sum of 16 4th Powers

## Theorem

Let $m \in \Z$ be an integer of the form $31 \times 16^n$ for $n \in \Z_{\ge 0}$.

Then in order to express $m$ as the sum of fourth powers, you need $16$ of them.

## Proof

We have:

- $31 \times 16^n = \paren {2^{n + 1} }^4 + 15 \times \paren {2^n}^4$

so every integer of the form $31 \times 16^n$ for $n \in \Z_{\ge 0}$ can be expressed as the sum of 16 fourth powers.

Now we show that we cannot use less than $16$ fourth powers.

Observe that for an even number $2 k$:

\(\ds \paren {2 k}^4\) | \(=\) | \(\ds 16 k^4\) | ||||||||||||

\(\ds \) | \(\equiv\) | \(\ds 0\) | \(\ds \pmod {16}\) |

For an odd number $2 k + 1$:

\(\ds \paren {2 k + 1}^4\) | \(=\) | \(\ds 16 k^4 + 32 k^3 + 24 k^2 + 8 k + 1\) | Binomial Theorem | |||||||||||

\(\ds \) | \(\equiv\) | \(\ds 8 k \paren {k + 1} + 1\) | \(\ds \pmod {16}\) | |||||||||||

\(\ds \) | \(\equiv\) | \(\ds 1\) | \(\ds \pmod {16}\) | since $k \paren {k + 1}$ is even |

It is obvious that for $n = 0$, $31$ requires $16$ fourth powers to express:

- $31 = 2^4 + 15 \times 1^4$

Aiming for a contradiction, suppose for some $n > 0$, $31 \times 16^n$ requires less than $16$ fourth powers to express.

Let $m$ be the smallest of those $n$.

Suppose $x$ of the summands are odd, where $x < 16$.

By the above, we must have:

- $31 \times 16^m \equiv x \pmod {16}$

Since $31 \times 16^m$ is divisible by $16$, $x = 0$.

Hence each summand is even.

Dividing each summand by $2$ gives a representation of $31 \times 16^{m - 1}$ as a sum of less than $16$ fourth powers.

This contradicts the minimality condition on $m$.

Hence the result by Proof by Contradiction.

$\blacksquare$